someone give me a quick overview of how to solve congruences? say 3x=24 mod blah blah you get it...I would really appreciate it...I already know how but I am double checking my knowledge base!

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- Mar 31st 2008, 06:58 PM #1

- Mar 31st 2008, 07:20 PM #2

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Say $\displaystyle 3x\equiv 24 (\bmod 60)$. Note that $\displaystyle \gcd(3,60)=3$ and $\displaystyle 3|24$, so the congruence is solvable and has exactly $\displaystyle 3$ different solutions (up to congruence). If you can find one solution $\displaystyle x_0$, the others are $\displaystyle x_0 + \frac{60}{3}$ and $\displaystyle x_0 + \frac{60}{3}\cdot 2$. So it just remains to find $\displaystyle x_0$. So for example, $\displaystyle x_0=8$ works. Now you can find everything else.